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What is the derivative?

  1. Feb 27, 2008 #1
    Hello there

    I know that derivative of a function is the tangent to it's curve and that is the limit of the function as H goes to zero but I still can not understand what it means?

    For example let's say F(x)=x^2 and F(x) is a function of a fighter jet plane speed, I know that if I take the slope of the secant line that passes through two points of the fighter jet speed, I'd get the average speed between two points, my question is whose value is the tangent? I read that it is the instaneous rate of change at the point itself rather than the average rate between two point, why it is not considered to be an average speed between two points and the second one is zero. what is the difference between X^2 and it's derivative 2x?

    Thanks
     
  2. jcsd
  3. Feb 27, 2008 #2
    I think that the best way to describe a derivative is to say that it is the slope of the curve at a specific point. I am sure you are familiar with a slope of a straight line, which is just the difference in y between two points on it divided by the difference in x between the two points. However the slope of a curve is not constant, it varies as a function of x. Calculus finds how the close varies by assuming that if two poins are veery close (aka 0 apart), then the path along the curve between them is just a straight line. This means that we can use the same old method to find the slope at that point. This is called the tangent and is the exact slope. If you go any farther than 0 apart, then you just get the average slope, which is called the secant. That is a at least the mathematical way to look at it.

    I am sure you also heard of slope being considered as the amount y changes for every change in x. So for example, let's say you know that y=f(x) where y is the displacement at time x. You also know that the velocity is given by change in displacement / change in time, so basically the slope of y=f(x). Now let's say that the displacement is modeled by y=x^2 and you want to know the velocity at time x=2. Since this is not a linear function, you can't just use the pick-any-two-points method to find the slope at x=2. You have to pick two points that are infinitely close to x, such that they are both x. And using calculus you know that the deriative of y=x^2 (which is dy/dx = 2x) will give you the slope at any arbitary point x. So the instantaneous velocity at time x=2 is dy/dx=2(2)=4.

    That is basically what a deriative is. If you want more detail on a part of my explanation, just ask. :)

    Btw, if F(x) is a function of speed, then the secant should give you the average acceleration. I think you wanted to say that F(x) is the function of displacement vs time.
     
  4. Feb 28, 2008 #3
    so a derivative is instaneous rate of change because the slope touches one point of the curve while the secant line touches couple points but the strange thing, if you subtract y2-y1 over x2-x1 by plugging values you will get average value while if you use algebriac manipulation, you get the slope of the tangent which has a different value than the first one.

    I have a question about Derivatives

    Take a look at this picture
    [​IMG]

    When you are taking the difference in values of y2-y1 over x2-x1, you get the difference over the difference which is the slope of the secant, when you take the slope closer and closer until you get close to zero, the slope of the tangent (( the brown line in the picture )) has the same length of the slope of secant and the slope of the tangent has Y2,Y1,X2,X1 but Y2 and X2 doesn't touches the curve of the function, it is outside the curve but does the derivative gives the value of the difference of the Y2-Y1 over X2-X1. I wanted to see if my I am thinking of is true but it needs precise graphing ,I hope you understand my question and thanks for your help man.
     
  5. Feb 28, 2008 #4
    I think you need to be a bit clearer in your question.

    "When you are taking the difference in values of y2-y1 over x2-x1, you get the difference over the difference which is the slope of the secant" --You are finding the difference in Vertical change, and the difference of horizontal change. And yes it is the slope of the secant provided that (x2, y2) is a second point on your curve.

    "The slope of the tangent has the same length of the slope of secant" --This makes no sense, slopes do not have "lengths".

    "I wanted to see if my I am thinking of is true but it needs precise graphing" --This makes no sense because of grammar.

    "but does the derivative gives the value of the difference of the Y2-Y1 over X2-X1" -- Yes, but the derivative is a limit, so think of (x2, y2) being infinitely close to (x1, y1).
     
  6. Feb 28, 2008 #5
    [​IMG]

    By the way, the Y2 of the point is supposed to be Y1 but I mistakenly written Y2.

    Many Math websites when trying to draw the derivative, they draw a secant line passing through two points and exeeding them and when they draw another one closer and closer, they deliberately draw lines that has the same length while they mean the difference between two points, I thought the derivative as the change of Y's over change of X's and I uploaded a picture that describes what I meant using the cartesian coordinate

    I wanted to see if what I was thinking of was true or not but I needed precise graphing because I could not draw a precise quadratic function.

    So a derivative means instaneous rate of change and a limit where the change in X gets arbitrarily close to zero
     
    Last edited: Feb 28, 2008
  7. Feb 28, 2008 #6

    CompuChip

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    You might also want to refer to this thread :smile:
     
  8. Feb 28, 2008 #7
    Thanks, your post is very informative.
     
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